MHB Likelihood Ratio Test for Common Variance from Two Normal Distribution Samples

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The discussion revolves around constructing a likelihood ratio test (LRT) for the common variance of two normal distributions under the null hypothesis that the variance equals a specified value. The likelihood functions for the two hypotheses are derived, leading to the calculation of the likelihood ratio and its transformation into a chi-squared distribution form. The poster questions the correctness of their approach compared to a textbook solution that presents a simpler chi-squared test without detailing the LRT steps. Clarifications are sought on the differences between the LRT and standard hypothesis tests, particularly regarding when to prefer one method over the other. The conversation highlights the complexity of LRT derivations compared to traditional methods.
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$\newcommand{\szdp}[1]{\!\left(#1\right)}
\newcommand{\szdb}[1]{\!\left[#1\right]}$
Problem Statement: Let $S_1^2$ and $S_2^2$ denote, respectively, the variances of independent random samples of sizes $n$ and $m$ selected from normal distributions with means $\mu_1$ and $\mu_2$ and common variance $\sigma^2.$ If $\mu_1$ and $\mu_2$ are unknown, construct a likelihood ratio test of $H_0: \sigma^2=\sigma_0^2$ against $H_a:\sigma^2=\sigma_a^2,$ assuming that $\sigma_a^2>\sigma_0^2.$

Note 1: This is Problem 10.89 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Sheaffer.

Note 2: This is cross-posted here.

My Work So Far: Let $X_1, X_2,\dots,X_n$ be the sample from the normal distribution with mean $\mu_1,$ and let $Y_1, Y_2,\dots,Y_m$ be the sample from the normal
distribution with mean $\mu_2.$ The likelihood is
\begin{align*}
L(\mu_1,\mu_2,\sigma^2)
=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\sigma^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\sigma^2}\szdp{\sum_{i=1}^n(x_i-\mu_1)^2
+\sum_{i=1}^m(y_i-\mu_2)^2}}.
\end{align*}
We obtain $L\big(\hat{\Omega}_0\big)$ by replacing $\sigma^2$ with $\sigma_0^2$ and $\mu_1$ with $\overline{x}$ and $\mu_2$ with $\overline{y}:$
\begin{align*}
L\big(\hat{\Omega}_0\big)
=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\sigma_0^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\sigma_0^2}\szdp{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}}.
\end{align*}
The MLE for the common variance in exactly this scenario (but with switched $m$ and $n$) is:
$$\hat\sigma^2=\frac{1}{m+n}\szdb{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}.$$
So this estimator plugged into the likelihood yields
\begin{align*}
L\big(\hat{\Omega}\big)
&=\szdp{\frac{1}{\sqrt{2\pi}}}^{\!\!(m+n)}
\szdp{\frac{1}{\hat\sigma^2}}^{\!\!(m+n)/2}
\exp\szdb{-\frac{1}{2\hat\sigma^2}\szdp{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}}.
\end{align*}
It follows that the ratio is
\begin{align*}
\lambda
&=\frac{L\big(\hat{\Omega}_0\big)}{L\big(\hat{\Omega}\big)}\\
&=\szdp{\frac{\hat\sigma^2}{\sigma_0^2}}^{\!\!(m+n)/2}
\exp\szdb{\frac{(\sigma_0^2-\hat\sigma^2)(m+n)}{2\sigma_0^2}}.\\
-2\ln(\lambda)
&=(m+n)\szdb{\frac{\hat\sigma^2}{\sigma_0^2}
-\ln\szdp{\frac{\hat\sigma^2}{\sigma_0^2}}-1}.
\end{align*}
Now the function $f(x)=x-\ln(x)-1$ first decreases, then increases. It has a global minimum of $0$ at $x=1.$ Note also that the original inequality becomes:
\begin{align*}
\lambda&<k\\
2\ln(\lambda)&<2\ln(k)\\
-2\ln(\lambda)&>k'.
\end{align*}
As the test is for $\sigma_a^2>\sigma_0^2,$ we will expect the estimator $\hat\sigma^2>\sigma_0^2.$ We can, evidently, use Theorem 10.2 and claim that $-2\ln(\lambda)$ is $\chi^2$ distributed with d.o.f. $1-0.$ So we reject $H_0$ when
$$(m+n)\szdb{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{(m+n)\sigma_0^2}
-\ln\szdp{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{(m+n)\sigma_0^2}}-1}
>\chi^2_{\alpha},$$
or
$$\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{\sigma_0^2}
-\ln\szdp{\frac{\sum_{i=1}^n(x_i-\overline{x})^2
+\sum_{i=1}^m(y_i-\overline{y})^2}{\sigma_0^2}}-(m+n)
>\chi^2_{\alpha}.$$

My Questions:

1. Is my answer correct?
2. My answer is not the book's answer. The book's answer is simply that
$$\chi^2=\frac{(n-1)S_1^2+(m-1)S_2^2}{\sigma_0^2}$$
has a $\chi_{(n+m-2)}^2$ distribution under $H_0,$ and that we reject $H_0$ if $\chi^2>\chi_a^2.$ How is this a likelihood ratio test? It's not evident that they went through any of the steps of forming the likelihood ratio with all the necessary optimizations. Their estimator is not the MLE for $\sigma^2,$ is it?
 
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Sorry for the necropost. Why/when would we choose to conduct a LRT over a standard hypothesis test?
 
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